Finding Moles and Atoms Question. Kinda Tough! (check my work)?

Question

Question: the mineral galena is composed of lead(II) sulfide and has an average density of 7.46 g/cm^3. (a) how many moles of lead(II) sulfide are in 1.00 ft^3 of galena? (b) how many lead atoms are in 1.00 dm^3 of galena?

My approach:

(a) 7.46g FeS x (10^6 cm^3/ 35.51 ft^3) x (1 mol/ 88 g FeS) = 2400.8 mol/1 ft^3 = 2400.8 mol FeS ---My Answer!

(b) 7.46 g/ 1cm^3 x (1000 cm^3/1 dm^3) = 7460 g/ 1 dm^3 x (1mol/88 g FeS) x (1 mol Fe atom/1mol) x (6.022e23 atoms/1mol) = 5.11e25 Fe Atoms/dm^3 --- my answer!

If you see anything wrong PLEASE tell me what's wrong by correcting it. appreciate it!

Answer 1

Kudos on attempting to work this out BEFORE posting the question. It demonstrates that you are thinking about the problem. Good job.

Galena is Lead Sulfide (PbS). One of the assumptions you have to make for this problem is that it doesn’t form hydrates. That would thrown off all your calcs. It doesn’t form hydrates.

The problem really is converting a cubic foot to cm^3. Convert feet to inches first 1 cuft = 12^3 cubic inches= 1728 cubic inches. Then convert cubic inches to cubic cm. 1728 x (2.54 cm/inch)^3 = 1728 x ( 16.4 cm^3/inch^3) = 28,300 cm^3

Now the weight of galena. 28,300 cm^3 x 7.46 g/ cm^3 = 211,000 g of galena.

Now the moles. 211,000g galena / 239.3 g galena per mole of galena = 883 moles of galena.

Last there is a 1 to 1 ratio of lead to Galena, so your answer is 883 moles of lead in 1 cubic foot of Galena.

BTW did you notice that you were working the problem with Iron Sulfide? That’s fool’s gold (pyrite), not galena!

Answer 2

I have asked the same question 4 times, and didn't get a proper answer